(sending to all - @Home is a non-functioning address) - Jay
"@Home" wrote:
> I had the following to solve:
>
> 51% of all domestic cars being shipped have power windows. If a lot contains
> five such cars:
>
> a. what is probability that only one has power windows?
> b. what is probability that at least one has power windows?
>
> I solved each of these problems in two ways, one using std probability
> theory and one by using a binomial distribution. I seemingly had no problem
> w/part b., but in part a. the probability theory did not seem to produce the
> correct answer. I have listed these below. What is wrong w/the probability
> equation listed below? Also is my answer to part b. correct?
>
> a. Randomly Draw Five Samples (Cars)
>
> Independent Events Only 1 w/Power Windows
>
> P{Only 1 Power} = P (Power) x P (NotPower) x P (NotPower) x P
> (NotPower) x P (NotPower)
> 0.51 0.49 0.49 0.49 0.49 =
Don't forget, you listed only 1 way to get 1 PW (power window) and 4 not.
There are 5 wys you could get this result, if you don't count the order (which
the question doesn't include). C(5,1) = 5!/(4!*1!) = 5.
So: 0.51*0.49*0.49 * 0.49 * 0.49 * 5 = = 0.14700
> Also Solve Using BINOMDIST Function in Excel
> n 5
> ? 0.51 Success - PW
> x 1
> p(x) 0.14700
>
> b. At least 1 w/Power Windows
> P {At Least 1} = 1 - P {0}
>
> P {0} = P (NotPower) x P (NotPower) x P (NotPower) x P (NotPower) x
> P (NotPower)
> 0.49 0.49 0.49 0.49 0.49
>
> Prob 0 0.02825
>
> 1 - 0.02825
> At least 1 0.97175
In this one, all the outcomes are alike, so there is no combination effect.
> Also Solve Using BINOMDIST Function in Excel ~ 97%
> n 5
> ? 0.49 Success - No Power
> x 0
> p(x) 0.02825
>
> 1 - 0.028247525
> 97%
So you got it! Or nearly so.
Cheers,
Jay
--
Jay Warner
Principal Scientist
Warner Consulting, Inc.
4444 North Green Bay Road
Racine, WI 53404-1216
USA
Ph: (262) 634-9100
FAX: (262) 681-1133
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